blocks|key|1710562|text|fooFunc()将首先执行，然后是bazFunc()，最后是barFunc()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1710563|我们都同意，在barFunc()进行任何操作之前，必须先执行fooFunc()。|1710564|鉴于bazFunc()仅在barFunc()需要其参数时才会被调用，因此这将在fooFunc()之后发生是顺理成章的。|1710565|entityMap^0|0|9|I|9|V|9|0|7|9|U|9|0|2|9|D|9|13|9|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|K|$]]

<code>fooFunc()</code> will execute first, followed by <code>bazFunc()</code> and finally <code>barFunc()</code>

We can all agree that <code>fooFunc()</code> must execute before <code>barFunc()</code> has anything to operate on.

Given that <code>bazFunc()</code> will only be called when <code>barFunc()</code> needs it's parameters, it stands to reason that this would happen after <code>fooFunc()</code>.

blocks|key|1710607|text|先是fooFunc，然后是bazFunc，最后是barFunc|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1710608|下面是一些演示它的代码：|1710609|public+class+OrderJava+{
++public+static+void+main(String[]+args)+{
++++fooFunc().barFunc(bazFunc());
++}

++public+static+Bar+fooFunc()+{
++++System.out.println("I+am+fooFunc!");
++++return+new+Bar();
++}

++public+static+class+Bar+{
++++public+void+barFunc(Object+o)+{
++++++System.out.println("I+am+barFunc!");
++++}
++}

++public+static+Object+bazFunc()+{
++++System.out.println("I+am+bazFunc!");

++++return+null;
++}
}|code-block|syntax|javascript|1710610|此代码的输出为：|1710611|I+am+fooFunc!
I+am+bazFunc!
I+am+barFunc!|1710612|entityMap^0|2|7|D|7|O|7|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|11|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|R|$]]

First <code>fooFunc</code>, then <code>bazFunc</code>, and last <code>barFunc</code>

Here's some code that demonstrates it:

<pre><code>public class OrderJava {
 public static void main(String[] args) {
 fooFunc().barFunc(bazFunc());
 }

 public static Bar fooFunc() {
 System.out.println("I am fooFunc!");
 return new Bar();
 }

 public static class Bar {
 public void barFunc(Object o) {
 System.out.println("I am barFunc!");
 }
 }

 public static Object bazFunc() {
 System.out.println("I am bazFunc!");

 return null;
 }
}
</code></pre>

The output of this code is:

<pre><code>I am fooFunc!
I am bazFunc!
I am barFunc!
</code></pre>

blocks|key|1710725|text|这方面的文档是15.12.4.+Run-time+Evaluation+of+Method+Invocation+|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1710726|它说“在运行时，方法调用需要五个步骤。第一，可以计算目标引用。第二，计算参数表达式。第三，检查要调用的方法的可访问性。第四，定位要执行的方法的实际代码。第五，创建新的激活帧，必要时执行同步，并将控制转移到方法代码。”|1710727|在本例中，fooFunc()是作为计算目标引用的一部分调用的，而bazFunc()是参数表达式之一，因此必须首先调用fooFunc()。|style|CODE|1710728|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.12.4^0|7|1E|0|0|0|5|9|W|9|1M|9|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@$A|S|B|T|1|U]]|C|$]]|$1|D|3|E|5|6|7|V|8|@]|9|@]|C|$]]|$1|F|3|G|5|6|7|W|8|@$A|X|B|Y|H|I]|$A|Z|B|10|H|I]|$A|11|B|12|H|I]]|9|@]|C|$]]|$1|J|3|-4|5|6|7|13|8|@]|9|@]|C|$]]]|K|$L|$5|M|N|O|C|$P|Q]]]]

The documentation for this is <a href="http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.12.4" rel="noreferrer">15.12.4. Run-time Evaluation of Method Invocation </a>

It says "At run-time, method invocation requires five steps. First, a target reference may be computed. Second, the argument expressions are evaluated. Third, the accessibility of the method to be invoked is checked. Fourth, the actual code for the method to be executed is located. Fifth, a new activation frame is created, synchronization is performed if necessary, and control is transferred to the method code."

In the example, <code>fooFunc()</code> is called as part of computing the target reference, and <code>bazFunc()</code> is one of the argument expressions, so <code>fooFunc()</code> must be called first.

In the statement:

<pre><code>fooFunc().barFunc(bazFunc());
</code></pre>

<code>barFunc()</code> can obviously not execute until both <code>bazFunc()</code> and <code>fooFunc()</code> have completed.

But is the order of execution of <code>fooFunc()</code> and <code>bazFunc()</code> guaranteed?

Related (but different!) question: <a href="https://stackoverflow.com/questions/2201688/order-of-execution-of-parameters-guarantees-in-java">Order of execution of parameters guarantees in Java?</a>

Order of execution of methods describing an instance and an argument in Java?

在声明中：fooFunc().barFunc(bazFunc());显然，只有在bazFunc()和fooFunc()都完成之后，barFunc()才能执行。但是fooFunc()和bazFunc()的执行顺序有保证吗？相关(但不同！)问题：

问Java中描述实例和参数的方法的执行顺序？
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问Java中描述实例和参数的方法的执行顺序？EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问Java中描述实例和参数的方法的执行顺序？
EN